HDU 1017 A Mathematical Curiosity【看懂题意+穷举法】
浏览数：20 /
时间：2015年06月11日
//2014.10.17 01:19
//题意：
//先输入一个数N，然后分块输入，每块输入每次2个数，n，m，直到n,m同时为零时
//结束，当a和b满足题目要求时那么这对a和b就是一组
//注意：
//每一块的输出中间有一个回车
A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27177 Accepted Submission(s): 8651
Problem Description
Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.
Output
For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.
Sample Input
1 10 1 20 3 30 4 0 0
Sample Output
Case 1: 2 Case 2: 4 Case 3: 5
Source
#include<stdio.h>
int main()
{
int N;
int n,m;
int a,b;
int cas;
scanf("%d",&N);
while(N--)
{
cas=1;
while(scanf("%d%d",&n,&m),n||m)
{
int count=0;
for(a = 1; a < n; a++)//穷举法
{
for(b = a + 1; b < n; b++)
{
if((a*a + b*b + m) % (a * b) == 0)
count++;
}
}
printf("Case %d: %d\n",cas++,count);
}
if(N)
printf("\n");
}
return 0;
}
郑重声明：本站内容如果来自互联网及其他传播媒体，其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享，并不代表本站赞同其观点和对其真实性负责，也不构成任何其他建议。
